\(\int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx\) [645]

Optimal result

Integrand size = 28, antiderivative size = 457 \[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=-\frac {8 e^2 (e f-3 d g) \sqrt {f+g x} \sqrt {a+c x^2}}{15 c g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}+\frac {2 \sqrt {-a} e \left (9 a e^2 g^2-c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {f+g x} \sqrt {1+\frac {c x^2}{a}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \sqrt {a+c x^2}}-\frac {2 \sqrt {-a} \left (a e^2 g^2 (7 e f-15 d g)-c \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )\right ) \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \sqrt {1+\frac {c x^2}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right ),-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {f+g x} \sqrt {a+c x^2}} \]

[Out]

-8/15*e^2*(-3*d*g+e*f)*(g*x+f)^(1/2)*(c*x^2+a)^(1/2)/c/g^2+2/5*e^2*(e*x+d)*(g*x+f)^(1/2)*(c*x^2+a)^(1/2)/c/g+2
/15*e*(9*a*e^2*g^2-c*(45*d^2*g^2-30*d*e*f*g+8*e^2*f^2))*EllipticE(1/2*(1-x*c^(1/2)/(-a)^(1/2))^(1/2)*2^(1/2),(
-2*a*g/(-a*g+f*(-a)^(1/2)*c^(1/2)))^(1/2))*(-a)^(1/2)*(g*x+f)^(1/2)*(1+c*x^2/a)^(1/2)/c^(3/2)/g^3/(c*x^2+a)^(1
/2)/((g*x+f)*c^(1/2)/(g*(-a)^(1/2)+f*c^(1/2)))^(1/2)-2/15*(a*e^2*g^2*(-15*d*g+7*e*f)-c*(-15*d^3*g^3+45*d^2*e*f
*g^2-30*d*e^2*f^2*g+8*e^3*f^3))*EllipticF(1/2*(1-x*c^(1/2)/(-a)^(1/2))^(1/2)*2^(1/2),(-2*a*g/(-a*g+f*(-a)^(1/2
)*c^(1/2)))^(1/2))*(-a)^(1/2)*(1+c*x^2/a)^(1/2)*((g*x+f)*c^(1/2)/(g*(-a)^(1/2)+f*c^(1/2)))^(1/2)/c^(3/2)/g^3/(
g*x+f)^(1/2)/(c*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {945, 1668, 858, 733, 435, 430} \[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\frac {2 \sqrt {-a} e \sqrt {\frac {c x^2}{a}+1} \sqrt {f+g x} \left (9 a e^2 g^2-c \left (45 d^2 g^2-30 d e f g+8 e^2 f^2\right )\right ) E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}}}-\frac {2 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}} \left (a e^2 g^2 (7 e f-15 d g)-c \left (-15 d^3 g^3+45 d^2 e f g^2-30 d e^2 f^2 g+8 e^3 f^3\right )\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right ),-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {a+c x^2} \sqrt {f+g x}}-\frac {8 e^2 \sqrt {a+c x^2} \sqrt {f+g x} (e f-3 d g)}{15 c g^2}+\frac {2 e^2 \sqrt {a+c x^2} (d+e x) \sqrt {f+g x}}{5 c g} \]

[In]

Int[(d + e*x)^3/(Sqrt[f + g*x]*Sqrt[a + c*x^2]),x]

[Out]

(-8*e^2*(e*f - 3*d*g)*Sqrt[f + g*x]*Sqrt[a + c*x^2])/(15*c*g^2) + (2*e^2*(d + e*x)*Sqrt[f + g*x]*Sqrt[a + c*x^
2])/(5*c*g) + (2*Sqrt[-a]*e*(9*a*e^2*g^2 - c*(8*e^2*f^2 - 30*d*e*f*g + 45*d^2*g^2))*Sqrt[f + g*x]*Sqrt[1 + (c*
x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(Sqrt[-a]*Sqrt[c]*f - a*g)])/(15*c^
(3/2)*g^3*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[a + c*x^2]) - (2*Sqrt[-a]*(a*e^2*g^2*(7*e*f
- 15*d*g) - c*(8*e^3*f^3 - 30*d*e^2*f^2*g + 45*d^2*e*f*g^2 - 15*d^3*g^3))*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f
+ Sqrt[-a]*g)]*Sqrt[1 + (c*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(Sqrt[-a
]*Sqrt[c]*f - a*g)])/(15*c^(3/2)*g^3*Sqrt[f + g*x]*Sqrt[a + c*x^2])

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2*a*Rt[-c/a, 2]*(d + e*x)^m*(Sqrt[1
+ c*(x^2/a)]/(c*Sqrt[a + c*x^2]*(c*((d + e*x)/(c*d - a*e*Rt[-c/a, 2])))^m)), Subst[Int[(1 + 2*a*e*Rt[-c/a, 2]*
(x^2/(c*d - a*e*Rt[-c/a, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-c/a, 2]*x)/2]], x] /; FreeQ[{a, c, d, e},
x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 945

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Simp[2*e^2*(d
+ e*x)^(m - 2)*Sqrt[f + g*x]*(Sqrt[a + c*x^2]/(c*g*(2*m - 1))), x] - Dist[1/(c*g*(2*m - 1)), Int[((d + e*x)^(
m - 3)/(Sqrt[f + g*x]*Sqrt[a + c*x^2]))*Simp[a*e^2*(d*g + 2*e*f*(m - 2)) - c*d^3*g*(2*m - 1) + e*(e*(a*e*g*(2*
m - 3)) + c*d*(2*e*f - 3*d*g*(2*m - 1)))*x + 2*e^2*(c*e*f - 3*c*d*g)*(m - 1)*x^2, x], x], x] /; FreeQ[{a, c, d
, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[2*m] && GeQ[m, 2]

Rule 1668

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}-\frac {\int \frac {-5 c d^3 g+a e^2 (2 e f+d g)+e \left (3 a e^2 g+c d (2 e f-15 d g)\right ) x+4 c e^2 (e f-3 d g) x^2}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx}{5 c g} \\ & = -\frac {8 e^2 (e f-3 d g) \sqrt {f+g x} \sqrt {a+c x^2}}{15 c g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}-\frac {2 \int \frac {-\frac {1}{2} c g^2 \left (15 c d^3 g-a e^2 (2 e f+15 d g)\right )+\frac {1}{2} c e g \left (9 a e^2 g^2-c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) x}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx}{15 c^2 g^3} \\ & = -\frac {8 e^2 (e f-3 d g) \sqrt {f+g x} \sqrt {a+c x^2}}{15 c g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}-\frac {\left (e \left (9 a e^2 g^2-c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right )\right ) \int \frac {\sqrt {f+g x}}{\sqrt {a+c x^2}} \, dx}{15 c g^3}+\frac {\left (a e^2 g^2 (7 e f-15 d g)-c \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )\right ) \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx}{15 c g^3} \\ & = -\frac {8 e^2 (e f-3 d g) \sqrt {f+g x} \sqrt {a+c x^2}}{15 c g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}-\frac {\left (2 a e \left (9 a e^2 g^2-c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {f+g x} \sqrt {1+\frac {c x^2}{a}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {2 a \sqrt {c} g x^2}{\sqrt {-a} \left (c f-\frac {a \sqrt {c} g}{\sqrt {-a}}\right )}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )}{15 \sqrt {-a} c^{3/2} g^3 \sqrt {\frac {c (f+g x)}{c f-\frac {a \sqrt {c} g}{\sqrt {-a}}}} \sqrt {a+c x^2}}+\frac {\left (2 a \left (a e^2 g^2 (7 e f-15 d g)-c \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )\right ) \sqrt {\frac {c (f+g x)}{c f-\frac {a \sqrt {c} g}{\sqrt {-a}}}} \sqrt {1+\frac {c x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 a \sqrt {c} g x^2}{\sqrt {-a} \left (c f-\frac {a \sqrt {c} g}{\sqrt {-a}}\right )}}} \, dx,x,\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )}{15 \sqrt {-a} c^{3/2} g^3 \sqrt {f+g x} \sqrt {a+c x^2}} \\ & = -\frac {8 e^2 (e f-3 d g) \sqrt {f+g x} \sqrt {a+c x^2}}{15 c g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+c x^2}}{5 c g}+\frac {2 \sqrt {-a} e \left (9 a e^2 g^2-c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {f+g x} \sqrt {1+\frac {c x^2}{a}} E\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \sqrt {a+c x^2}}-\frac {2 \sqrt {-a} \left (a e^2 g^2 (7 e f-15 d g)-c \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )\right ) \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \sqrt {1+\frac {c x^2}{a}} F\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 c^{3/2} g^3 \sqrt {f+g x} \sqrt {a+c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 24.72 (sec) , antiderivative size = 619, normalized size of antiderivative = 1.35 \[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\frac {\sqrt {f+g x} \left (\frac {2 e^2 (-4 e f+15 d g+3 e g x) \left (a+c x^2\right )}{c g^2}+\frac {2 (f+g x) \left (\frac {e g^2 \sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}} \left (-9 a e^2 g^2+c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \left (a+c x^2\right )}{(f+g x)^2}+\frac {\sqrt {c} e \left (-i \sqrt {c} f+\sqrt {a} g\right ) \left (-9 a e^2 g^2+c \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {\frac {g \left (\frac {i \sqrt {a}}{\sqrt {c}}+x\right )}{f+g x}} \sqrt {-\frac {\frac {i \sqrt {a} g}{\sqrt {c}}-g x}{f+g x}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}{\sqrt {f+g x}}\right )|\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )}{\sqrt {f+g x}}+\frac {\sqrt {c} g \left (15 i c^{3/2} d^3 g^2+9 a^{3/2} e^3 g^2-i a \sqrt {c} e^2 g (2 e f+15 d g)+\sqrt {a} c e \left (-8 e^2 f^2+30 d e f g-45 d^2 g^2\right )\right ) \sqrt {\frac {g \left (\frac {i \sqrt {a}}{\sqrt {c}}+x\right )}{f+g x}} \sqrt {-\frac {\frac {i \sqrt {a} g}{\sqrt {c}}-g x}{f+g x}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}{\sqrt {f+g x}}\right ),\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )}{\sqrt {f+g x}}\right )}{c^2 g^4 \sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}\right )}{15 \sqrt {a+c x^2}} \]

[In]

Integrate[(d + e*x)^3/(Sqrt[f + g*x]*Sqrt[a + c*x^2]),x]

[Out]

(Sqrt[f + g*x]*((2*e^2*(-4*e*f + 15*d*g + 3*e*g*x)*(a + c*x^2))/(c*g^2) + (2*(f + g*x)*((e*g^2*Sqrt[-f - (I*Sq
rt[a]*g)/Sqrt[c]]*(-9*a*e^2*g^2 + c*(8*e^2*f^2 - 30*d*e*f*g + 45*d^2*g^2))*(a + c*x^2))/(f + g*x)^2 + (Sqrt[c]
*e*((-I)*Sqrt[c]*f + Sqrt[a]*g)*(-9*a*e^2*g^2 + c*(8*e^2*f^2 - 30*d*e*f*g + 45*d^2*g^2))*Sqrt[(g*((I*Sqrt[a])/
Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g)/Sqrt[c] - g*x)/(f + g*x))]*EllipticE[I*ArcSinh[Sqrt[-f - (I*Sqr
t[a]*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*Sqrt[a]*g)/(Sqrt[c]*f + I*Sqrt[a]*g)])/Sqrt[f + g*x] + (Sqrt[c
]*g*((15*I)*c^(3/2)*d^3*g^2 + 9*a^(3/2)*e^3*g^2 - I*a*Sqrt[c]*e^2*g*(2*e*f + 15*d*g) + Sqrt[a]*c*e*(-8*e^2*f^2
+ 30*d*e*f*g - 45*d^2*g^2))*Sqrt[(g*((I*Sqrt[a])/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g)/Sqrt[c] - g*x
)/(f + g*x))]*EllipticF[I*ArcSinh[Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*Sqrt[a]*g)/(
Sqrt[c]*f + I*Sqrt[a]*g)])/Sqrt[f + g*x]))/(c^2*g^4*Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]])))/(15*Sqrt[a + c*x^2])

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 711, normalized size of antiderivative = 1.56

method	result	size
elliptic	\(\frac {\sqrt {\left (g x +f \right ) \left (c \,x^{2}+a \right )}\, \left (\frac {2 e^{3} x \sqrt {c g \,x^{3}+c f \,x^{2}+a g x +f a}}{5 c g}+\frac {2 \left (3 d \,e^{2}-\frac {4 f \,e^{3}}{5 g}\right ) \sqrt {c g \,x^{3}+c f \,x^{2}+a g x +f a}}{3 c g}+\frac {2 \left (d^{3}-\frac {2 f a \,e^{3}}{5 c g}-\frac {a \left (3 d \,e^{2}-\frac {4 f \,e^{3}}{5 g}\right )}{3 c}\right ) \left (\frac {f}{g}-\frac {\sqrt {-a c}}{c}\right ) \sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x -\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x +\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}}\, F\left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}, \sqrt {\frac {-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\right )}{\sqrt {c g \,x^{3}+c f \,x^{2}+a g x +f a}}+\frac {2 \left (3 d^{2} e -\frac {3 e^{3} a}{5 c}-\frac {2 f \left (3 d \,e^{2}-\frac {4 f \,e^{3}}{5 g}\right )}{3 g}\right ) \left (\frac {f}{g}-\frac {\sqrt {-a c}}{c}\right ) \sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x -\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x +\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}}\, \left (\left (-\frac {f}{g}-\frac {\sqrt {-a c}}{c}\right ) E\left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}, \sqrt {\frac {-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\right )+\frac {\sqrt {-a c}\, F\left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}, \sqrt {\frac {-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\right )}{c}\right )}{\sqrt {c g \,x^{3}+c f \,x^{2}+a g x +f a}}\right )}{\sqrt {g x +f}\, \sqrt {c \,x^{2}+a}}\)	\(711\)
risch	\(\text {Expression too large to display}\)	\(1082\)
default	\(\text {Expression too large to display}\)	\(2950\)

[In]

int((e*x+d)^3/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

((g*x+f)*(c*x^2+a))^(1/2)/(g*x+f)^(1/2)/(c*x^2+a)^(1/2)*(2/5*e^3/c/g*x*(c*g*x^3+c*f*x^2+a*g*x+a*f)^(1/2)+2/3*(
3*d*e^2-4/5*f/g*e^3)/c/g*(c*g*x^3+c*f*x^2+a*g*x+a*f)^(1/2)+2*(d^3-2/5*f*a/c/g*e^3-1/3*a/c*(3*d*e^2-4/5*f/g*e^3
))*(f/g-(-a*c)^(1/2)/c)*((x+f/g)/(f/g-(-a*c)^(1/2)/c))^(1/2)*((x-(-a*c)^(1/2)/c)/(-f/g-(-a*c)^(1/2)/c))^(1/2)*
((x+(-a*c)^(1/2)/c)/(-f/g+(-a*c)^(1/2)/c))^(1/2)/(c*g*x^3+c*f*x^2+a*g*x+a*f)^(1/2)*EllipticF(((x+f/g)/(f/g-(-a
*c)^(1/2)/c))^(1/2),((-f/g+(-a*c)^(1/2)/c)/(-f/g-(-a*c)^(1/2)/c))^(1/2))+2*(3*d^2*e-3/5*e^3/c*a-2/3*f/g*(3*d*e
^2-4/5*f/g*e^3))*(f/g-(-a*c)^(1/2)/c)*((x+f/g)/(f/g-(-a*c)^(1/2)/c))^(1/2)*((x-(-a*c)^(1/2)/c)/(-f/g-(-a*c)^(1
/2)/c))^(1/2)*((x+(-a*c)^(1/2)/c)/(-f/g+(-a*c)^(1/2)/c))^(1/2)/(c*g*x^3+c*f*x^2+a*g*x+a*f)^(1/2)*((-f/g-(-a*c)
^(1/2)/c)*EllipticE(((x+f/g)/(f/g-(-a*c)^(1/2)/c))^(1/2),((-f/g+(-a*c)^(1/2)/c)/(-f/g-(-a*c)^(1/2)/c))^(1/2))+
(-a*c)^(1/2)/c*EllipticF(((x+f/g)/(f/g-(-a*c)^(1/2)/c))^(1/2),((-f/g+(-a*c)^(1/2)/c)/(-f/g-(-a*c)^(1/2)/c))^(1
/2))))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.70 \[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=-\frac {2 \, {\left ({\left (8 \, c e^{3} f^{3} - 30 \, c d e^{2} f^{2} g + 3 \, {\left (15 \, c d^{2} e - a e^{3}\right )} f g^{2} - 45 \, {\left (c d^{3} - a d e^{2}\right )} g^{3}\right )} \sqrt {c g} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (c f^{2} - 3 \, a g^{2}\right )}}{3 \, c g^{2}}, -\frac {8 \, {\left (c f^{3} + 9 \, a f g^{2}\right )}}{27 \, c g^{3}}, \frac {3 \, g x + f}{3 \, g}\right ) + 3 \, {\left (8 \, c e^{3} f^{2} g - 30 \, c d e^{2} f g^{2} + 9 \, {\left (5 \, c d^{2} e - a e^{3}\right )} g^{3}\right )} \sqrt {c g} {\rm weierstrassZeta}\left (\frac {4 \, {\left (c f^{2} - 3 \, a g^{2}\right )}}{3 \, c g^{2}}, -\frac {8 \, {\left (c f^{3} + 9 \, a f g^{2}\right )}}{27 \, c g^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (c f^{2} - 3 \, a g^{2}\right )}}{3 \, c g^{2}}, -\frac {8 \, {\left (c f^{3} + 9 \, a f g^{2}\right )}}{27 \, c g^{3}}, \frac {3 \, g x + f}{3 \, g}\right )\right ) - 3 \, {\left (3 \, c e^{3} g^{3} x - 4 \, c e^{3} f g^{2} + 15 \, c d e^{2} g^{3}\right )} \sqrt {c x^{2} + a} \sqrt {g x + f}\right )}}{45 \, c^{2} g^{4}} \]

[In]

integrate((e*x+d)^3/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

-2/45*((8*c*e^3*f^3 - 30*c*d*e^2*f^2*g + 3*(15*c*d^2*e - a*e^3)*f*g^2 - 45*(c*d^3 - a*d*e^2)*g^3)*sqrt(c*g)*we
ierstrassPInverse(4/3*(c*f^2 - 3*a*g^2)/(c*g^2), -8/27*(c*f^3 + 9*a*f*g^2)/(c*g^3), 1/3*(3*g*x + f)/g) + 3*(8*
c*e^3*f^2*g - 30*c*d*e^2*f*g^2 + 9*(5*c*d^2*e - a*e^3)*g^3)*sqrt(c*g)*weierstrassZeta(4/3*(c*f^2 - 3*a*g^2)/(c
*g^2), -8/27*(c*f^3 + 9*a*f*g^2)/(c*g^3), weierstrassPInverse(4/3*(c*f^2 - 3*a*g^2)/(c*g^2), -8/27*(c*f^3 + 9*
a*f*g^2)/(c*g^3), 1/3*(3*g*x + f)/g)) - 3*(3*c*e^3*g^3*x - 4*c*e^3*f*g^2 + 15*c*d*e^2*g^3)*sqrt(c*x^2 + a)*sqr
t(g*x + f))/(c^2*g^4)

Sympy [F]

\[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{\sqrt {a + c x^{2}} \sqrt {f + g x}}\, dx \]

[In]

integrate((e*x+d)**3/(g*x+f)**(1/2)/(c*x**2+a)**(1/2),x)

[Out]

Integral((d + e*x)**3/(sqrt(a + c*x**2)*sqrt(f + g*x)), x)

Maxima [F]

\[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{2} + a} \sqrt {g x + f}} \,d x } \]

[In]

integrate((e*x+d)^3/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^3/(sqrt(c*x^2 + a)*sqrt(g*x + f)), x)

Giac [F]

\[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{2} + a} \sqrt {g x + f}} \,d x } \]

[In]

integrate((e*x+d)^3/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^3/(sqrt(c*x^2 + a)*sqrt(g*x + f)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^3}{\sqrt {f+g\,x}\,\sqrt {c\,x^2+a}} \,d x \]

[In]

int((d + e*x)^3/((f + g*x)^(1/2)*(a + c*x^2)^(1/2)),x)

[Out]

int((d + e*x)^3/((f + g*x)^(1/2)*(a + c*x^2)^(1/2)), x)